Integrating the Yukawa Potential: Understanding the 3rd Step

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1st step: go to "spherical coordinates" and integrating out the [itex]\phi[/itex]... [itex]u = \cos \theta[/itex]
2nd step: Integrating wrt to [itex]u[/itex], will give a difference of [itex]e^{i~something} - e^{-i~something} \propto \sin (something)[/itex]. In fact I wouldn't ever write it in terms of sin...
3rd step: uses that he has two integrals one with the exponential with + and the other with the exponential with - ... then changes the integration variable of the one from k to -k, and gets this result.
4th step: factorizes the denominator.
5th step and then final: solves the (complex) integral

If you want to see a step in more details, you can ask for a specific one.
 
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I actually get it with your points! Thank you very much!
 
ChrisVer said:
1st step: go to "spherical coordinates" and integrating out the [itex]\phi[/itex]... [itex]u = \cos \theta[/itex]
2nd step: Integrating wrt to [itex]u[/itex], will give a difference of [itex]e^{i~something} - e^{-i~something} \propto \sin (something)[/itex]. In fact I wouldn't ever write it in terms of sin...
3rd step: uses that he has two integrals one with the exponential with + and the other with the exponential with - ... then changes the integration variable of the one from k to -k, and gets this result.
4th step: factorizes the denominator.
5th step and then final: solves the (complex) integral

If you want to see a step in more details, you can ask for a specific one.
hello sir,
i can't get the 3rd step. is change the integration variable, change the value from k to -k in that term, including the dk -> d(-k)?
 
Muh. Fauzi M. said:
i can't get the 3rd step. is change the integration variable, change the value from k to -k in that term, including the dk -> d(-k)?

like everytime you change your integration variable from [itex]x[/itex] to [itex]y(x)[/itex] what has to change is of course the differential, the integrand and the limits of the integral.